# Integration of an inverse sqrt composite function

by Hendrick
Tags: composite, function, integration, inverse, sqrt
 P: 43 1. The problem statement, all variables and given/known data “Geologist A” at the bottom of a cave signals to his colleague “Geologist B” at the surface by pushing a 11.0 kg box of samples from side to side. This causes a transverse wave to propagate up the 77.0 m rope. The total mass of the rope is 14.0 kg. Take g = 9.8 m/s². How long does it take for the wave to travel from the bottom of the cave to the surface?[Hint: Find an analytic expression v(z) for the wave speed as a function of distance. Then use the fact that at any given point on the rope the time dt taken to travel a small distance dz is given by: dt=dz/v(z). Then integrate to obtain the total travel time. ] 2. Relevant equations u = mR/z T(z) = u.z.g + mB.g v(z) = (T(z)/u)^(1/2) dt=dz/v(z) z = the length of the rope = L (used for integrating) 3. The attempt at a solution v(z) = (T(z)/u)^(1/2) v(z) = ((mR/z).z.g + mB.g/(mR/z))^(1/2) v(z) = ((mR.g + mB.g)/(mR/z))^(1/2) v(z) = ([(mR.g)/(mR/z)] + [(mB.g)/(mR/z)])^(1/2) v(z) = ([(mR.g.z)/mR] + [(mB.g.z)/mR])^(1/2) v(z) = ([mR.g.z] + [(mB.g.z)/mR])^(1/2) dt=dz/v(z) dt=dz/([mR.g.z] + [(mB.g.z)/mR])^(1/2) Integration: ....f L t= | (1/([mR.g.z] + [(mB.g.z)/mR])^(1/2)).dz ....j 0 ....f L t= | 2.([mR.g.z^2/2] + [(mB.g.z^2)/mR.2])^(1/2).dz ....j 0 I think I integrated it properly but when substituted the values mB = mass of box mR = mass of rope g = 9.8 ms^2 z = 77.0 m I didn't get the correct answer of t = 2.52s
 Sci Advisor HW Helper Thanks P: 25,228 'u' in your solution is supposed to be the mass density of the rope. It's not mR/z. The rope doesn't have a variable density, it's mR/(total length of rope), a constant. Nice problem presentation, by the way.
P: 43
 Quote by Dick 'u' in your solution is supposed to be the mass density of the rope. It's not mR/z. The rope doesn't have a variable density, it's mR/(length of rope). Nice problem presentation, by the way.
Hi Dick, z is the length of the rope (77.0m as in the problem) it was just the letter they used in the formula sheet so I carried it forth.

How was my integration? I don't think I know how to integrate nested functions (I assume it's something like the reverse of the chain-rule?).

HW Helper
Thanks
P: 25,228
Integration of an inverse sqrt composite function

 Quote by Hendrick Hi Dick, z is the length of the rope (77.0m as in the problem) it was just the letter they used in the formula sheet so I carried it forth. How was my integration? I don't think I know how to integrate nested functions (I assume it's something like the reverse of the chain-rule?).
z in your problem is the variable indicating length along the rope. I mean that u=14 kg/(77 m). It doesn't have the variable of integration in it. Until that gets fixed there isn't any point in discussing the integral.
 Sci Advisor HW Helper Thanks P: 25,228 In your notation u=mR/L not mR/z.
P: 43
 Quote by Dick z in your problem is the variable indicating length along the rope. I mean that u=14 kg/(77 m). It doesn't have the variable of integration in it. Until that gets fixed there isn't any point in discussing the integral.
Oh, ok. Thanks for pointing that out :)

So:-

u = mR/L = mR/77
T(z) = u.z.g + mB.g
v(z) = (T(z)/u)^(1/2)
dt=dz/v(z)

z = the length of the rope = L (used for integrating)

3. The attempt at a solution
v(z) = (T(z)/u)^(1/2)
v(z) = ((u.z.g + mB.g)/u)^(1/2)
v(z) = ([(u.z.g)/u] + [(mB.g)/u])^(1/2)
v(z) = ([g.z] + [(mB.g)/u])^(1/2)

dt=dz/v(z)
dt=dz/([g.z] + [(mB.g)/u])^(1/2)

Integration:

....f L
t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).dz
....j 0

....f L
t= | 2.([g.z] + [(mB.g)/u])^(1/2).(g.z^2)/2
....j 0

Where:
mB = mass of box
mR = mass of rope
g = 9.8 ms^2
z = 77.0 m

Is this correct?
 Sci Advisor HW Helper Thanks P: 25,228 Your first integral looks just fine. I don't know how you got from there to the second one. The usual way to do an integration like this is to do a change of variable. Let v=gz+mBg/u.
P: 43
 Quote by Dick Your first integral looks just fine. I don't know how you got from there to the second one. The usual way to do an integration like this is to do a change of variable. Let v=gz+mBg/u.
Ok, I was trying to use the chain rule lol ><

So:-

Integration:

....f L
t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).dz
....j 0

....f L
t= | 2.([(g.z^2)/2] + [(mB.g.z)/u] + C)^(1/2)
....j 0

Where:
mB = mass of box
mR = mass of rope
g = 9.8 ms^2
z = 77.0 m

How am I doing?
 Sci Advisor HW Helper Thanks P: 25,228 Not too good. You're coming up with some pretty bizarre integration rules which aren't in the book. You have to do the change of variable thing. Eg to integrate 1/(a+bz)^(1/2) I would say v=(a+bz), so dv=b*dz. This turns the integral into 1/v^(1/2)*dv*(1/b). Now it's just integrating v^(-1/2). Does that sound familiar?
P: 43
 Quote by Dick Not too good. You're coming up with some pretty bizarre integration rules which aren't in the book. You have to do the change of variable thing. Eg to integrate 1/(a+bz)^(1/2) I would say v=(a+bz), so dv=b*dz. This turns the integral into 1/v^(1/2)*dv*(1/b). Now it's just integrating v^(-1/2). Does that sound familiar?
Unfortunately not very familiar at all. I haven't really dealt with composite integrals. But I'll try:

a = g.z
b = (mB.g)/u
1/(a+bz)^(1/2)

v=(a+bz), so dv=b*dz.
This turns the integral into 1/v^(1/2)*dv*(1/b).
Integrating v^(-1/2).

So:-

Integration:

....f L
t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).dz
....j 0

....f L
t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).[(mB.g)/u]dz.(1/[(mB.g)/u])
....j 0

....f L
t= | 2.(([g.z] + [(mB.g)/u])^(1/2)).[(mB.g)/u].(1/[(mB.g)/u])
....j 0

Where:
mB = mass of box
mR = mass of rope
g = 9.8 ms^2
z = 77.0 m
 Sci Advisor HW Helper Thanks P: 25,228 Take a break and clear your head. While your at it look back at integration by substitution in a calc text. I can roughly see what you are trying to do - but you still seem to be trying to do some kind of a chain rule. And the (1/b) factor in the example becomes (1/g) in the problem, right? Do you see where it's coming from? And after the integration is done and the dz is gone you should also drop the integral sign - it looks pretty confusing otherwise.

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